Arrange the following in order of increasing number of unpaired electrons in them i.…

1. i, iv, ii, iii
2. iv, i, iii, ii
3. iv, i, ii, ii
4. i, iv, iii, ii

Solution

(i) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}:-$ $\mathrm{Fe}$ is in +3 state and $\mathrm{CN}^{-}$is a strong-field ligand. Thus, $\mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^5$ or
Thus, there is only one unpaired electron. (ii) $\left[\mathrm{MnCl}_6\right]^{3-}$ :- $\mathrm{Mn}$ is in +3 state and $\mathrm{Cl}^{-}$is a weak-field ligand. Thus, $\mathrm{Mn}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^4$ or
Thus, there are four unpaired electrons. (iii) $\left[\mathrm{FeF}_6\right]^{3-}:-$ $\mathrm{Fe}$ is in +3 state and $\mathrm{F}^{-}$is a weak-field ligand. Thus, Fe will be $[\mathrm{Ar}] 3 \mathrm{~d}^5$ with five unpaired electron.
(iv) $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}$ :- $\mathrm{Co}$ is in +3 state here and $\mathrm{NH}_3$ is a strong-field ligand. Thus, $\mathrm{Co}$ will be $[\mathrm{Ar}] 3 \mathrm{~d}^6$ with zero unpaired electrons.
Thus, the increasing order of the number of unpaired electrons will be :- (iv) $ < $ (i) $ < $ (ii) < (iii)

Asked in: AP EAMCET 2023 (15 May Shift 2)

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